Chapter 3: Q. 3.39 (page 121)

In Problem $2.32$ you computed the entropy of an ideal monatomic gas that lives in a two-dimensional universe. Take partial derivatives with respect to $U, A$ , and N to determine the temperature, pressure, and chemical potential of this gas. (In two dimensions, pressure is defined as force per unit length.) Simplify your results as much as possible, and explain whether they make sense.

Short Answer

Expert verified

$\begin{aligned} T = \frac{U}{N k} \\ P = \frac{N k T}{A} \\ μ = - k T \ln (\frac{V}{N} {(\frac{2 m m k T}{h^{2}})}^{\frac{3 / 2}{2}}) \end{aligned}$

Step by step solution

Given Information

The entropy of an ideal gas is,

$S = N k [\ln (\frac{2 π m A U}{(N h)^{2}}) + 2]$

Calculation

To get the temperature, we partial differentiate with respect to U,

$\begin{matrix} \frac{1}{T} = {(\frac{\partial S}{\partial U})}_{A, N} \\ \frac{1}{T} = \frac{\partial}{\partial U} [N k [\ln (\frac{2 m m A U}{(N h)^{2}}) + 2]] \\ = N k [\frac{(N h)^{2}}{2 x m A U} \frac{2 nmA}{(N h)^{2}}] \\ = \frac{N k}{U} \\ T = \frac{U}{N k} \end{matrix}$

Partial differentiating with respect to $A$ , we get the pressure as,

$\begin{matrix} P = T (\frac{\partial S}{\partial A}) U, N \\ = T \frac{\partial}{\partial A} [N k [\ln (\frac{2 m n U U}{(N h)^{2}}) + 2]] \\ = N k T [\frac{(N h)^{2}}{2 π n A U} \times \frac{2 m m U}{(N h)^{2}}] \\ = \frac{N k T}{A} \end{matrix}$

Chemical Potential

Partial differentiating the entropy with respect to N, we get the chemical potential as,

$\begin{matrix} μ = - T (\frac{\partial S}{d N}) U A \\ = - T \frac{\partial}{d N} [N k [\ln (\frac{2 π m A U}{(N h)^{2}}) + 2]] \\ = - k T [\ln (\frac{2 π m A U}{(N h)^{2}}) + 2] - k T N \frac{(N h)^{2}}{2 π m A U} \times \frac{2 π n A U}{h^{2}} \times \frac{- 2}{N^{3}} \\ = - k T [\ln (\frac{2 m m A U}{(N h)^{2}}) + 2] + 2 k T \\ = - k T [\ln (\frac{2 m m A U}{(N h)^{2}}) + 2 - 2] \\ = - k T [\ln (\frac{2 m m A U}{(N h)^{2}})] \\ = - k T [\ln (\frac{2 m m A (N k T)}{(N h)^{2}})] \\ = - k T \ln {(\frac{V}{N} (\frac{2 π m L T}{N^{2}}))}^{3 / 2}) \end{matrix}$

Conclusion

The temperature, pressure and entropy are given by:

$\begin{aligned} T = \frac{U}{N k} \\ P = \frac{N k T}{A} \\ μ = - k T \ln (\frac{V}{N} {(\frac{2 π m k T}{h^{2}})}^{\frac{3}{2}}) \end{aligned}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Given Information

Calculation

Chemical Potential

Conclusion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Space Physics

Atoms and Radioactivity

Quantum Physics

Conservation of Energy and Momentum

Energy Physics

Rotational Dynamics

Study anywhere. Anytime. Across all devices.